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zzz [600]
3 years ago
6

Is aluminum Pure or Impure? ​

Chemistry
1 answer:
mr_godi [17]3 years ago
6 0

Answer:

The aluminium is impure

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Kaitlyn said that the metal sodium Na and the non metal neon Ne form the ionic compound NaNe. Using what you know about groups a
Maslowich
Based on what I know about groups and valence electrons I support kaitlyn’s statement due to the fact that an ionic bond is a bond between a metal and nonmetal.
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How are cistrans isomers used for night vision?
Paladinen [302]

Qual das alternativas a seguir é uma mistura homogênea?

Escolha uma:a. Mercúrio metálico e água. b. Gelo e água líquida. c. Areia e carvão em pó. d. Nitrogênio gasoso e vapor d’água. e. Poeira e ar atmosférico.
7 0
4 years ago
CH3COOH  CH3COO– + H+
Oxana [17]

(a)

pH = 4.77

; (b)

[

H

3

O

+

]

=

1.00

×

10

-4

l

mol/dm

3

; (c)

[

A

-

]

=

0.16 mol⋅dm

-3

Explanation:

(a) pH of aspirin solution

Let's write the chemical equation as

m

m

m

m

m

m

m

m

l

HA

m

+

m

H

2

O

⇌

H

3

O

+

m

+

m

l

A

-

I/mol⋅dm

-3

:

m

m

0.05

m

m

m

m

m

m

m

m

l

0

m

m

m

m

m

l

l

0

C/mol⋅dm

-3

:

m

m

l

-

x

m

m

m

m

m

m

m

m

+

x

m

l

m

m

m

l

+

x

E/mol⋅dm

-3

:

m

0.05 -

l

x

m

m

m

m

m

m

m

l

x

m

m

x

m

m

m

x

K

a

=

[

H

3

O

+

]

[

A

-

]

[

HA

]

=

x

2

0.05 -

l

x

=

3.27

×

10

-4

Check for negligibility

0.05

3.27

×

10

-4

=

153

<

400

∴

x

is not less than 5 % of the initial concentration of

[

HA

]

.

We cannot ignore it in comparison with 0.05, so we must solve a quadratic.

Then

x

2

0.05

−

x

=

3.27

×

10

-4

x

2

=

3.27

×

10

-4

(

0.05

−

x

)

=

1.635

×

10

-5

−

3.27

×

10

-4

x

x

2

+

3.27

×

10

-4

x

−

1.635

×

10

-5

=

0

x

=

1.68

×

10

-5

[

H

3

O

+

]

=

x

l

mol/L

=

1.68

×

10

-5

l

mol/L

pH

=

-log

[

H

3

O

+

]

=

-log

(

1.68

×

10

-5

)

=

4.77

(b)

[

H

3

O

+

]

at pH 4

[

H

3

O

+

]

=

10

-pH

l

mol/L

=

1.00

×

10

-4

l

mol/L

(c) Concentration of

A

-

in the buffer

We can now use the Henderson-Hasselbalch equation to calculate the

[

A

-

]

.

pH

=

p

K

a

+

log

(

[

A

-

]

[

HA

]

)

4.00

=

−

log

(

3.27

×

10

-4

)

+

log

(

[

A

-

]

0.05

)

=

3.49

+

log

(

[

A

-

]

0.05

)

log

(

[

A

-

]

0.05

)

=

4.00 - 3.49

=

0.51

[

A

-

]

0.05

=

10

0.51

=

3.24

[

A

-

]

=

0.05

×

3.24

=

0.16

The concentration of

A

-

in the buffer is 0.16 mol/L.

hope this helps :)

6 0
2 years ago
At 25°C and 1.0 atm, 0.84 g of a gas dissolves in 2.00 L of water. What mass of the gas dissolves
UNO [17]

The mass of the gas dissolved in 1.00 L of water at 25°C and 3.0 atm is equal to 1.26 grams.

<h3>How to determine the mass of the gas dissolved?</h3>

In order to determine the mass of the gas dissolved, we would the calculate the new (final) solubility of this gas by applying this formula:

S₁P₂ = S₂P₁

Making S₂ the subject of formula, we have:

S₂ = (S₁P₂)/P₁

S₂ = (0.42 × 3.0)/1.0

S₂ = 1.26 g/L.

Now, we can determine the mass:

Mass = solubility × volume

Mass = 1.26 × 1.00

Mass = 1.26 grams.

Read more on solubility here: brainly.com/question/3006391

#SPJ1

5 0
2 years ago
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Scorpion4ik [409]
Because your better prepared and get a better understanding of the lab your doing?
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